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The solubility product of Al(OH)3 is 1×10-36.Calculate the solubility of Al(OH)3. |
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Answer» Al(OH)3 \(\leftrightharpoons\) Al+3(aq)+ + 3OH-(aq) using law of mass action K = \(\frac{[Al^{+3}][OH^-]^3}{[Al(OH^-)_3]}\) ⇒ K = equilibrium constant \(\because\) [al(OH)_3] = constant \(\therefore\) Rearranging above equation. K x [Al(OH)3] = [Al+3] [OH-]3 ⇒ Ksp = [Al+3] [OH-]3 where Ksp = [S] [3S]3 = s x 27S3 Ksp = 27s4 We have given, Ksp = 1 x 10-36 \(\therefore\) S4 = \(\frac1{27}\times10^{-36}\) = 3.7 x 10-38 S4 = 370 x 10-40 S = 4.39 x 10-10 g/L Hence, solubility of Al(OH)3 will be 4.39 x 10-10 g/L |
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